817g of an unknowm gas occupies 1.25l at 27.0°c and a pressure of 765 mmhg. is the gas most likely methane(ch4), ethane(c2h6), propane(c3h8), or butane(c4h10)? show work.

Use the equation pv=nrt

101991.60949Pa(0.00125m^3)=n(8.31)(300°K)
n=(101991.60949Pa*0.00125m^3)/(8.31*300°K)
n=0.0511moles

You have grams, which is .817, and you have moles. The next step is molar mass.

.817g/.0511moles=15.988g/mol

Now, calculate the molar mass of each gas.
CH4 = 16.043g/mol
C2H6 = 30.07g/mol
C3H8 = etc
C4H10 = etc

The closest answer is your CH4 (methane), so that is your answer. Hope this helps!

P.S. If you wouldn't mind, if you could mark this answer as so I can keep moving up levels! Thanks :)

I’m not sure what your trying to say here.. can you clarify again?